:解方程组 {(x+y)/2 +(x-y)/3 =6{ 3(x+y)-2(x-y)=12

问题描述:

:解方程组
{(x+y)/2 +(x-y)/3 =6
{ 3(x+y)-2(x-y)=12

x=3,y=9

{(x+y)/2 +(x-y)/3 =6
{ 3(x+y)-2(x-y)=12
设x+y=a x-y=b
3a+2b=6
3a-2b=12
6a=18
a=3
b=-3/2
∴x+y=3
x-y=-3/2
2x=3/2
x=3/4
2y=9/2
y=9/4
∴x=3/4
y=9/4

令x+y=u
x-y=v
u/2+v/3=6
3u-2v=12
3u+2v=36
相加得
6u=48
u=8
从而
v=6

x+y=8
x-y=6
所以
x=7
y=1

3(x+y)+2(x-y)=36
3(x+y)-2(x-y)=12
令x+y=a,x-y=b
3a+2b=36(1)
3a-2b=12(2)
(1)+(2) 6a=48
a=8
(1)-(2) 4b=24
b=6
x+y=8
x-y=6
2x=14
x=7
y=1

可以用换元法:
令x+y=m,x-y=n
则原方程组等价于{m/2+n/3=6 {m=8
{3m-2n=12 解得:{n=6
即{x+y=8 {x=7
{x-y=6 解得:{y=1